Two Loop RC Circuit 2 1 2 3 4 5 6 A circuit is constructed with four resistors,

Question

Two Loop RC Circuit 2

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A circuit is constructed with four resistors, one capacitor, one battery and a switch as shown. The values for the resistors are: R1 = R2 = 35 , R3 = 76 and R4 = 119 . The capacitance is C = 66 F and the battery voltage is V = 12 V. The positive terminal of the battery is indicated with a + sign.

1)

The switch has been open for a long time when at time t = 0, the switch is closed. What is I4(0), the magnitude of the current through the resistor R4 just after the switch is closed?

A

2)

What is Q(), the charge on the capacitor after the switch has been closed for a very long time?

C

3)

After the switch has been closed for a very long time, it is then opened. What is Q(topen), the charge on the capacitor at a time topen = 546 s after the switch was opened?

C

4)

What is IC,max(closed), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is closed? A positive value for the current is defined to be in the direction of the arrow shown.

A

5)

What is IC,max(open), the current that flows through the capacitor whose magnitude is maximum during the time when the switch is open? A positive value for the current is defined to be in the direction of the arrow shown.

A

Explanation / Answer

Knowns:

        V = 12V

        R = 35

        R = 35

        R = 76

        R = 119

        C = 66F

(1) You should consider the capacitor to be in the "discharged" state just prior to t=0. Therefore, the voltage across it is 0V. This causes R to be effectively in parallel with R. That parallel combination is in series then with R and R. Given that, you can compute the current.

        Itotal = I = I = V (R+R+(R || R))

        I = 12V (35+119+(35 || 76))

      I = 12V (35+119+23.963)

        I 0.0674 A

(2) The capacitor, at t=, has no more currents leaving or arriving as everything has reached equilibrium by then. R's current is also therefore zero and so R doesn't drop any voltage. Without current through R and C, the capacitor can be simply removed without affecting the circuit (or treated as an resistance.) And so can R. This leaves only R, R, and R to worry about. First get the currents:

        Itotal = I = I = I = V (R+R+R)

        I = 12V (35+76+119)

        I 0.05217 A

The voltage drop across R will be the voltage across C:

        V = I R = 0.05217A 76 3.965V

Q = CVs = 66uF * 3.965V = 261.68 uC

(3) The only components to worry about with the switch open are C, R and R. The starting voltage on C is V = 3.965V, from above. You should know that:

        V(t) = V e^(-t ((R+R)C))

        V(t=546s) = 3.965V e^(-546s ((35+76)66F))

        V(546s) = 3.965V e^(-546s ((35+76)66F))

        V(546s) 3.68V

Q = CV = 66uF * 3.68V = 242.9 uC

(4)

Since R and R are in series, replace them with a single (R1+R4) = 154 resistor (short R and set R=154.) Now consider the (-) terminal of V to be the GND (0V) node. Convert V, the new R, and R into their Thevenin equivalent:

            Vth = VR (R+R) = 12V76 (154+76) 3.965V

            Rth = RR (R+R) 50.89

Rth is now in series with R and Vc=0V at t=0, so all of V appears across this new series pair. The current is:

            Ic(max) = Vth (Rth+R) = 3.965V (50.89+35) 0.046 A

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